160k views
2 votes
Pattern question i guess ? I need part c) working please thankiew :))

Pattern question i guess ? I need part c) working please thankiew :))-example-1
User Lyla
by
7.5k points

1 Answer

6 votes
From "Series 1" we know, that:


1+2+3+\ldots+n=(n(n+1))/(2)

and from "Series 2":


1^3+2^3+3^3+\ldots+n^3=(1+2+3+\ldots+n)^2

c)


3^3+6^3+9^3+\ldots+30^3=(1\cdot3)^3+(2\cdot3)^3+(3\cdot3)^3+\ldots+(10\cdot3)^3=\\\\=1^3\cdot3^3+2^3\cdot3^3+3^3\cdot3^3+\ldots+10^3\cdot3^3=\\\\=(1^3+2^3+3^3+\ldots+10^3)\cdot3^3\,\stackrel{\text{Series 2}}{=}\,(1+2+3+\ldots+10)^2\cdot27=\\\\\stackrel{\text{Series1}}{=}\,\left((10\cdot11)/(2)\right)^2\cdot27=55^2\cdot27=3025\cdot27=\boxed{81675}
User Dymk
by
6.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.